Problem: $\begin{aligned} &f(a)=-\dfrac{1}{4}(a+8) \\\\ &g(b)=\dfrac{2}{3}b+1 \end{aligned}$ $g(f(6))=$
Explanation: When evaluating composite functions, we work our way inside out. To evaluate $g(f(6))$, let's first evaluate $f(6)$. Then we'll plug that result into $g$ to find our answer. Let's evaluate $f({6})$. $\begin{aligned}f(a)&=-\dfrac{1}{4}(a+8)\\\\ f({6})&=-\dfrac{1}{4}\left({6}+8\right)~~~~~~~~~~\text{Plug in }a={6}\\\\ &=-\dfrac{1}{4}(14)\\\\ &={-\dfrac{7}{2}}\end{aligned}$ We now know that $g(f({6}))$ is the same as $g\left({-\dfrac{7}{2}}\right)$ because $f({6}) = {-\dfrac{7}{2}}$. Let's evaluate $g\left({-\dfrac{7}{2}}\right)$. $\begin{aligned}g(b)&=\dfrac{2}{3}b+1\\\\ g\left({{-\dfrac{7}{2}}}\right)&=\dfrac{2}{3}\left({-\dfrac{7}{2}}\right)+1~~~~~~~~~~\text{Plug in }b={-\dfrac{7}{2}}\\\\ &=-\dfrac{7}{3}+1\\\\ &=-\dfrac{4}{3}\\\\\end{aligned}$ The answer: $g(f(6))=-\dfrac{4}{3}$